on the exponential sum over r-free integers

Acta Math. Hungar.

90 (3) (2001), 219{230.

ON THE EXPONENTIAL SUM OVER r-FREE

INTEGERS

A. BALOG and I. Z. RUZSA, corresponding member of the Academy (Budapest)�

Abstract. We determine, up to a constant factor, the L1 mean of the ex-

ponential sum formed with the r-free integers. This improves earlier results ofBr�udern, Granville, Perelli, Vaughan and Wooley. As an application, we improvethe known bound for the L1 norm of the exponential sum de�ned with the M�obiusfunction.

1. Introduction

Br�udern, Granville, Perelli, Vaughan and Wooley [3] studied bounds forthe L1 mean of the exponential sum formed with the r-free integers, that is,integers not divisible by any r-th power > 1. They established that the L1

mean of this exponential sum is substantially smaller than the square rootof the L2 mean, but still a positive power of the length of the exponentialsum.

Let r = 2 be �xed and let an be the characteristic function of the r-freeintegers, that is

an =

�1; if n is r-free,

0; otherwise,a0 = 0; a�n = an:

Let N be a su�ciently large integer and de�ne

g1(�) =

NXn=1

ane(n�); g2(�) =Xjnj5N

ane(n�);

g3(�) =Xjnj5N

�1�

jnjN

�ane(n�):

The main result of [2] states that

(1:1) N1

2r �Z 1

0

��g1(�)�� d�� N1

r+1+"

� Research supported in part by OTKA grant T 017433 and T 025617.

0236{5294/1/$ 5.00 c 2001 Akad�emiai Kiad�o, Budapest

220 A. BALOG and I. Z. RUZSA

for all positive ". The authors also expressed their opinion that the lowerbound in (1.1) re ects the true order of magnitude.

In this paper we will show that the upper bound in (1.1) is almost bestpossible. We determine the precise order of magnitude.

Theorem 1. We have

(1:2) N1

r+1 �Z 1

0

��g1(�)�� d�� N1

r+1 :

The lower bound in (1.2) follows from the corresponding statement forg3(�), namely

Theorem 2. We have

(1:3) N1

r+1 �Z 1

0

��g3(�)�� d�� N1

r+1 :

On the other hand, the upper bound in (1.3) follows from that in (1.2),as we will immediately explain.

As the de�nition of g3(�) suggests, our basic tool is Fej�er's kernel(1:4)

0 5 F (�) =Xjnj5N

�1�

jnjN

�e(n�) =

sin2 �N�

N sin2 �� min

�N;

1

Nk�k2

�:

More generally we will consider for any integers 1 5 K and 0 5 N

(1:5)X

jnj5N+K

min

�1;N +K � jnj

K

�e(n�)

=sin�(2N +K)� sin�K�

K sin2 �� min

�N +K;

1

k�k;

1

Kk�k2

�:

The theorems are connected by the following relations:

��g2(�)�� 5 2��g1(�)�� ; g3(�) =

Z 1

0

F (�� )g2(�) d�:

UsingR 10F (�) d� = 1 we arrive at

(1:6)

Z 1

0

��g3(�)�� d� 5Z 1

0

Z 1

0

F (�� )��g2(�)�� d�d� 5 2

Z 1

0

��g1(�)�� d�:Acta Mathematica Hungarica 90, 2001

ON THE EXPONENTIAL SUM OVER r-FREE INTEGERS 221

Therefore it is su�cient to prove the upper bound of Theorem 1, which isdone in Section 2, and the lower bound of Theorem 2, which is the content ofSection 3. In Section 4 we give an application to the exponential sum formedwith the M�obius function.

2. Proof of the upper bound

We recall that an =P

drjn�(d). Let 1 5 y < z be any real numbers. For

n 6= 0 we de�ne cn = cn(y; z) as a middle part of the above sum, namely

(2:1) cn =Xdrjn

y5d<z

�(d);

and we put c0 = 0.

Lemma 1. For any 1 5 K < N and 1 5 y < z we have

(2:2)X

N�K<n5N

jcnj2 � Ky1�r +N1=r log3 z:

Proof. We have(2:3) X

N�K<n5N

jcnj2 5X

N�K<n5N

� Xdrjn

y5d<z

1

�2

=X

y5d1<z

Xy5d2<z

XN�K<n5Ndr1jn; dr

2jn

1:

The inner sum is bounded by K=[d1; d2]r + 1 whenever [d1; d2] 5 N1=r, and

empty otherwise. Thus, when it is not empty we have that 1 5 N1=r=[d1; d2].Therefore we can continue (2.3) as

(2:4) 5X

y5d1<z

Xy5d2<z

K

[d1; d2]r +

N1=r

[d1; d2]

!

5Xd1=y

Xd2=y

K

dr1dr2

Xtj(d1;d2)

tr +Xd1<z

Xd2<z

N1=r

d1d2

Xtj(d1;d2)

t

= KXt

tr� Xd=y; tjd

1

dr

�2

+N1=rXt<z

t

� Xd<z; tjd

1

d

�2

Acta Mathematica Hungarica 90, 2001


5 KXt

t�r

min

1;

�t

y

�r�1!!2

+N1=rXt<z

1

t

�Xd<z

1

d

�2

� KXt5y

tr�2

y2r�2+K

Xt>y

1

tr+N1=r log3 z � Ky1�r +N1=r log3 z:

This concludes the proof of the lemma.

We continue with the upper estimate of Theorem 1.Next note that for any integers 0 5 N , 1 5 d 5 N +K and arbitrary M

we have from (1.5) that

(2:5)X

jnj5N+K

n�M(d)

min

�1;N +K � jnj

K

�e(n�)

= e(M�)X

jmj5(N+K)=d

min

1;

�N+Kd

�� jmj

�K

d

�

!e(md�) +O(1)

� min

�N +K

d;

1

kd�k;

d

Kkd�k2

�:

Let D = N1=(r+1), and let de�ne S by 2S�1 < D 5 2S . Note that S �logN . We decompose an as follows

(2:6) an =

SXs=1

cn(D2�s;D2�s+1) + cn(D;N1=r):

We have

(2:7)

Z 1

0

��NXn=1

ane(n�)

�� d�

5

SXs=1

Z 1

0

��NXn=1

cn(D2�s;D2�s+1)e(n�)

�� d�+

Z 1

0

��NXn=1

cn(D;N1=r)e(n�)

�� d�:The Cauchy{Schwarz inequality, Parseval' identity and Lemma 1 imply that

(2:8)

Z 1

0

��NXn=1

cn(D;N1=r)e(n�)

�� d�Acta Mathematica Hungarica 90, 2001


5

sXn5N

��cn(D;N1=r)�� 2 �q

ND1�r +N1=r log3N � N1=(r+1):

For a �xed s we abbreviate cn(D2�s;D2�s+1) by cn and we also introduceM = [N=2] and K = N2�rs. We have

(2:9)

NXn=1

cne(n�) =X

jnj5M+K

min

�1;2M +K � jnj

K

�cM+ne

�(M + n)�

�

�X

2M<n52M+K

2M +K � jnjK

cne(n�)�Xn5K

K � jnjK

cne(�n�) +O(1):

The contribution of the second and third sums can be handled much thesame way as (2.8). This yields that

(2:10)

Z 1

0

�� X2M<n52M+K

2M +K � jnjK

cne(n�)

�� d� 5s X

2M<n52M+K

jcnj2

5

qK(D2�s)1�r +N1=r log3N � N1=(r+1)2�s=2 +N1=(2r) log3=2N;

and the same for the contribution of the last sum. As for the �rst sum weuse (2.5) and get

(2:11)

Z 1

0

�� Xjnj5M+K

min

�1;2M +K � jnj

K

�cM+ne

�(M + n)�

�� d�

5X

D2�s5d<D2�s+1

Z 1

0

��X

jnj5M+K

n��M(d)

min

�1;2M +K � jnj

K

�e�(M + n)�

�� d�

�X

D2�s5d<D2�s+1

Z 1

0

min

�N

d;

1

kd�k;

d

Kkd�k2

�d�

�X

D2�s5d<D2�s+1

Z 1

0

min

�N

d;

1

k�k;

d

Kk�k2

�d�

�X

D2�s5d<D2�s+1

�1 + log

N

K

�� s2�sD:



Adding all terms from (2.8), (2.10) and (2.11) we obtain

(2:12)

Z 1

0

��g1(�)�� d��

SXs=1

(s2�sD +N1=(r+1)2�s=2 +N1=(2r) log3=2N) +N1=(r+1) � N1=(r+1):

3. The lower bound

We retain the notation of cn(y; z) and D = N1=(r+1) and we put

(3:1) Fh(�) =Xjnj5Nhjn

�1�

jnjN

�e(n�) =

1

h

hXa=1

F��

a

h

�= 0:

We start with the decomposition

(3:2) g3(�) =Xjnj5N

�1�

jnjN

�ane(n�)

=Xjnj5N

�1�

jnjN

�cn(1;D)e(n�) +

Xjnj5N

�1�

jnjN

�cn(D;N

1=r)e(n�)

=Xd5D

�(d)Fdr (�)�Xd5D

�(d) +Xjnj5N

�1�

jnjN

�cn(D;N

1=r)e(n�):

(The second sum of the right-hand side compensates for the constant termsthat appear in Fh, while a0 = c0(x; y) = 0.) We denote the �rst sum in (3.2)by g4(�) and transform it to

(3:3) g4(�) =Xd5D

�(d)

dr

drXa=1

F��

a

dr

�=

=Xd5D

�(d)

dr

drXa=1

(a;dr) r-free

F��

a

dr

� Xm5D=d(m;d)=1

�(m)

mr=Xd5D

�(d)bdGd(�);



where

bd =X

m5D=d(m;d)=1

�(m)

mr

and

Gd(�) =1

dr

drXa=1

(a;dr) r-free

F��

a

dr

�= 0:

We can trivially estimate bd as

(3:4)1

35 1�

��2

6� 1

�5 1�

1Xd=2

d�r 5 bd 5 1 +

1Xd=2

d�r 5�2

65

5

3:

We are going to isolate a large part of the interval (0; 1) on which one termof g4(�) in (3.3) dominates. To this end we note that

(3:5) F (�) =N

2; whenever k�k 5

1

2N:

For any �xed d 5 D let us de�ne the set

(3:6) Xd =[

15a5dr

(a;dr) r-free

�a

dr�

1

2N;a

dr+

1

2N

�:

Note that the sets Xd are not necessarily disjoint for di�erent d 5 D. Forany � 2 Xd we have

(3:7) Gd(�) =N

2dr;

and we show that on a large part of Xd this term actually dominates. Wede�ne

(3:8) Yd =�� 2 Xd :

Xd05Dd0 6=d

Gd0(�) 5N

20dr

�:

The sets Yd are pairwise disjoint because for d 6= d0 � 2 Yd implies that

Gd(�) =N

2drand Gd0(�) 5

N

20dr



while � 2 Yd0 implies that

Gd0(�) =N

2d0rand Gd(�) 5

N

20d0r

which contradict each other. In case of � 2 Yd we have in view of (3.4) that

(3:9)��g4(�)�� = bdGd(�)�

Xd05Dd06=d

bd0Gd0(�) =N

12dr:

We have to estimate jYdj from below. First note that

jXdj =1

N

drX

a=1(a;dr) r-free

1 =dr

N

Ypjd

�1�

1

pr

�;

and therefore

(3:10)dr

N� jXdj 5

dr

N:

Next we estimate the size of Zd = Xd n Yd. If � 2 Zd then

Gd(�) =N

2dr; and

Xd05Dd0 6=d

Gd0(�) =N

20dr;

and this immediately implies that

(3:11)

ZZd

Gd(�)Xd05Dd0 6=d

Gd0(�) d� = jZdjN2

40d2r:

For the opposite estimate of the integral in (3.11) we prove a \quasi{orthogonality" property of the functions Gd(�), namely

Lemma 2. For d 6= d0 we have

Z 1

0

Gd(�)Gd0(�) d� � 1:



Proof. First we observe that

F (�)F (�) 5�F (�) + F (�)

�min

�F (�); F (�)

��F (�) + F (�)

�min

�N;

1

Nk�k2;

1

Nk�k2

�

5�F (�) + F (�)

�min

�N;

1

Nk�� k2

�

implying immediately that

(3:12)

Z 1

0

F (�)F (� + �) d� � min

�N;

1

Nk�k2

�:

By the de�nition of Gd(�) and by (3.12) we have

(3:13)

Z 1

0

Gd(�)Gd0 (�) d�

=1

drd0r

drXa=1

(a;dr) r-free

d0rX

a0=1(a0;d0r) r-free

Z 1

0

F��

a

dr

�F

��

a0

d0r

�d�

�1

drd0r

drXa=1

(a;dr) r-free

d0rX

a0=1(a0;d0

r

) r-free

min

N;

1

N a

dr� a0

d0r

2

!:

We will analyse the spacing behaviour of the rational numbers a=dr, where(a; dr) is r-free. We can write

(3:14)

adr � a0

d0r

= a d

0r

(dr ;d0r)� a0 d

r

(dr;d0r)

[dr; d0r]

= jmj[dr; d0r]

;

where m is the least absolute value residue of

(3:15) ad0r

(dr; d0r)� a0

dr

(dr; d0r)mod [dr; d0

r]:

m = 0 does not appear because d 6= d0 and (a; dr), (a0; d0r) are r-free. For

any other jmj 5 [dr; d0r]=2 we need that

ad0r

(dr; d0r)� m mod

dr

(dr; d0r);



which happens exactly for (dr; d0r) choices of a mod dr. Having a �xed, a0

is uniquely determined. In (3.13) we will use N to estimate the minimumwhenever jmj 5 [dr; d0

r]=N . We arrive at

Z 1

0

Gd(�)Gd0(�) d��1

[dr; d0r]

Xm5[dr;d0r]=2

min

N;

[dr; d0r]2

Nm2

!

�N

[dr; d0r]

Xm5[dr;d0r]=N

1 +[dr; d0

r]

N

Xm>[dr;d0r]=N

1

m2� 1:

The proof of Lemma 2 is therefore complete.

By means of Lemma 2 we can estimate (3.11) from above to get

ZZd

Gd(�)Xd05Dd0 6=d

Gd0(�) d� 5Xd05Dd0 6=d

Z 1

0

Gd(�)Gd0(�) d� � D

which together with (3.11) imply that

(3:16) jZdj �d2rD

N2:

Comparing this to (3.10) we can see that Zd is at most half of Xd, that is Ydis at least half of Xd whenever d is not too large. More precisely we chooseD0 = "D with a su�ciently small but positive ". We have

(3:17) jYdj �dr

N; if d 5 D0:

Finally we choose

(3:18) Y =[d5D0

Yd:

Note that by (3.10)

(3:19) jYj 5Xd5D0

jXdj 5Dr+1

0

N= "r+1:



From (3.9) and (3.17) we haveZY

��g4(�)�� d� = N

12

Xd5D0

jYdjdr

� "D;

and by (3.2), the Cauchy{Schwarz inequality, Lemma 1 and (3.19) we arriveat Z 1

0

��g3(�)�� d� =ZY

��g3(�)�� d�=

ZY

��g4(�)�� d�� jYjD �sjYj

Xjnj5N

��cn(D;N1=r)�� 2

� D("� "r+1 �p"r+1)� N

1

r+1 ;

if " > 0 is small enough. This completes the proof of Theorem 2.

4. An integral with the M�obius function

Improving a result of Balog and Perelli [1], in [2] we proved that

(4:1)

Z 1

0

��NXn=1

�(n)e(n�)

�� d�� N1=8

logN:

The results of this paper can be used to improve this estimate.

Theorem 3. Let bn be complex numbers satisfying

jbnj =�1 if n is r-free,

0 otherwise.

We have

(4:2)

Z 1

0

��NXn=1

bne(n�)

�� d�� N1=2(r+1):

In particular,

(4:3)

Z 1

0

��NXn=1

�(n)e(n�)

�� d�� N1=6:


230 A. BALOG and I. Z. RUZSA: ON THE EXPONENTIAL SUM OVER r-FREE INTEGERS

Proof. Write

h(�) =

NXn=1

bne(n�):

We have Z 1

0

h(�)h(� + �) d� =X

jbnj2e(n�) = g1(�):

HenceZ 1

0

��g1(�)�� d� 5Z 1

0

Z 1

0

��h(�)h(� + �)�� d� d� =

�Z 1

0

��h(�)�� d��2

;

and the claim follows from Theorem 1.

We remark that the method of [2], while giving a weaker bound, workedfor a wider class of coe�cients, the only restriction being that they vanishexcept for squarefree values. The method of [1] gave even weaker results anddid not yield such extensions, but it is the only one which works also for theLiouville function �. We are still unable to prove that

Z 1

0

��NXn=1

�(n)e(n�)

�� d�� N c

holds with some positive c.

References

[1] A. Balog and A. Perelli, On the L1 mean of the exponential sum formed with the

M�obius function, J. London Math. Soc., 57 (1998), 275{288.[2] A. Balog and I. Z. Ruzsa, A new lower bound for the L1 mean of the exponential sum

with the M�obius function, Bull. London Math. Soc., 31 (1999), 415{418.[3] J. Br�udern, A. Granville, A. Perelli, R. C. Vaughan and T. D. Wooley, On the ex-

ponential sum over k-free numbers, Philos. Trans. Roy. Soc. London, Ser. A,356 (1998), 739{761.

(Received March 10, 1999)

ALFR�ED R�ENYI MATHEMATICAL INSTITUTE

BUDAPEST 1364

P.O.BOX 127

E-MAIL: [email protected]

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on the exponential sum over r-free integers

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